Re: defining a closed set

On Sun, 26 Oct 2008 04:37:26 +0000, I wrote:

On Sat, 25 Oct 2008 18:38:14 -0700 (PDT), RichD
<r_delaney2001@xxxxxxxxx> wrote:

[...] intuitively, does
not a closed set consist of an open set, plus its
boundary?

[...] A regular open (or closed)
subset of a topological space divides the space up into three
disjoint subsets, in a particularly simple kind of way, rather
like the way a physical body seems to partition physical space
into its inside, its boundary, and its outside [...]

Let X be a topological space, and let F be any regular closed
subset of X. That is to say, let F be any subset of X such
that F = cl(int(F)).

Let U = int(F). Then int(cl(U)) = int(F) = U, i.e. U is a
regular open subset of X.

We could have started by letting U be any regular open subset
of X, i.e. any subset of X satisfying U = int(cl(U)): because
then, on defining F = cl(U), we have cl(int(F)) = cl(U) = F.

The interior operation on subsets of X, restricted to the regular
closed subsets of X, and the closure operation on subsets of X,
restricted to the regular open subsets of X, are mutually inverse
operations, thus:

cl
---->
{regular open subsets} {regular closed subsets}
<----
int

Either way, we have an open set U, and a closed set F, with
U a subset of F, F the closure of U, and U the interior of F.

Let B = F \ U. Then B is the boundary of F, and B is also the
boundary of U. F is the union of the disjoint sets U and B.

Let U' = X \ F = ext(U). Then U' is also a regular open set,
and U = ext(U'). The boundary of U' is also B, and the closure
of U', a regular closed set F', is the union of U' and B.

The whole space X is the union of the three disjoint subsets U,
B, and U', thus:

X
===============================
U B U'
----------------...------------
___________________
F _______________
F'

No set-theoretic or topological operation performed on any of
the sets in the tripartition of the space X takes you outside
the same tripartition of X.

I think this is probably the kind of picture you had in mind.

It works for regular open and closed subsets, but not for all
open or closed subsets. (For an indication of how messy things
can get, in the general case, look up the "Kuratowski closure
and complement problem", or some roughly equivalent phrase.)

Consideration of more than one regular open (or regular closed)
subset at the same time leads to the "regular open algebra" of
the space X.

(Take with a pinch of salt, in case I have got some details wrong.)

--
Angus Rodgers
In the kingdom of the blind, the one-eyed man will be
poked in the eye with a sharp stick.
.

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